If $\sum_{n = 1}^\infty {{a_n}}$ converges, then is $\sum_{n = 1}^\infty (1+a_n)^{-1}$ a convergent series? If $\sum\limits_{n = 1}^\infty  {{a_n}}$ is convergent (with ${a_n} > 0$, $\forall n\in\mathbb{Z}$), then is $\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{1 + {a_n}}}} \right)}$ is a convergent series?
I tried with the quotient criterion and comparison theorem no avail.
But I figured something was this:
$${\frac{1}{{1 + {a_n}}}}<1$$
but still I find it useful.
Any suggestions please.
 A: If $\sum a_n$ converges, then $a_n \to 0$, therefore $\frac{1}{1+a_n} \to 1$ and $\sum \frac{1}{1+a_n}$ diverges.
A: If the first series converges, then surely $a_n \rightarrow 0$. If that's true, can the terms in the second series also approach zero? If the second series is to converge, then they must. Said differently, if they don't, then the second series can't converge.
So if $\lim_{n \rightarrow \infty} a_n = 0$, what is $\lim_{n \rightarrow \infty} \frac{1}{1+a_n}$ ?
A: A more interesting situation is to consider $$\tag 1 \sum_{i\geqslant 1}\frac{a_i}{1+a_i}$$
If $a_i>0$, and $$\tag 2 \sum_{i\geqslant 1} a_i$$ exists, we have that $\dfrac{1}{1+a_i}\to 1$. Since this is the quotient of the $i$-th term of $(1)$ with that of $(2)$, this means that $(1)$ converges if $(2)$ does.  But if $(1)$ converges, from $\dfrac{a_i}{1+a_i}\to 0$ and $1+a_i>0$ we get that $a_i\to 0$. This gives again a quotient of $1$ and by comparison $(2)$ converges.  
A: If $\sum a_n <\infty$ then of course $\lim a_n =0$. Therefore $\lim \frac{1}{1+a_n} =1$, so second series does not satisfy  the term-test http://en.wikipedia.org/wiki/N-th_term_test
